Pre measure for an infinite product of measure spaces Let $\{(\Omega_k, \Sigma_k, P_k)\}_{k\geq 1}$ be a sequence of probability spaces. I am trying to prove the statement below in order to use it and get a pre measure and then use the Hahn kolomogrov theorem to get a probability measure on $\prod_{k=1}^{\infty} \Omega_k$. The statement I want to show is:
For every $k\geq 1$, let $\{E^{(k)}_j\}_{j\geq 1}$ be a sequence in $\Sigma_k$. It is given that:
1)  $\forall j\geq 1[ \,\,\,|\{k\geq 1:E^{(k)}_j\not= \Omega_k\}|<\infty \,\,\,\,]$
2) $\prod _{k=1}^{\infty} \Omega_k=\cup_{j=1}^{\infty}(\prod _{k=1}^{\infty} E^{(k)}_j)  $ and the sets in the collection $\{\prod _{k=1}^{\infty} E^{(k)}_j\}_{j\geq 1}$ are pairwise disjoint.
then $\sum_{j=1}^{\infty}(\prod _{k=1}^{\infty} P_k(E^{(k)}_j))=1  $.
I tried to prove it but only succeeded in proving special cases. If anyone can give me hints, that would be great.
Thank you a lot.
 A: First, let's introduce some notation.  Fix $n\geq 1$, and suppose $F_k\subseteq\Omega_k$ are measurable sets for each $k\geq n$ such that all but finitely many of them are $\Omega_k$.  Let $P^n(\prod_{k=n}^\infty F_k)=\prod P_k(F_k)$.  It is not hard to show that this gives rise to a finitely additive function on the Boolean algebra of finite unions of such products (you prove this in the course of constructing finite product measures).  Write $A_j=\prod_k E_j^{(k)}$; we wish to show that $\sum_j P^1(A_j)=1$.  It follows from finite additivity of $P^1$ and disjointness of the $A_j$ that $\sum_j P^1(A_j)\leq 1$.  So suppose for a contradiction that $\sum_j P^1(A_j)<1$.
Fix $x\in\Omega_1$, and write $A_j(x)=\prod_{k=2}^\infty E_j^{(k)}$ if $x\in E_j^{(1)}$, and $A_j(x)=\emptyset$ if $x\not\in E_j^{(1)}$.  Then the function $f_j(x)=P^2(A_j(x))$ is a measurable function on $\Omega_1$ for each $j$, with $\int_{\Omega_1} f_j=P^1(A_j)$.  Now consider the function $f(x)=\sum_j f_j(x)$.  Since $\sum_j P^1(A_j)<1$, $\int_{\Omega_1} f<1$, so there exists some $x_1\in\Omega_1$ such that $f(x_1)<1$, i.e. such that $\sum_j P^2(A_j(x_1))<1$.
Now repeat the argument of the previous paragraph, with the sets $A_j$ replaced by the sets $A_j(x_1)$ and $P^1$ replaced by $P^2$.  The argument above gives that there exists $x_2\in \Omega_2$ such that $\sum_j P^3(A_j(x_1,x_2))<1$ (where $A_j(x_1,x_2)$ is defined in the obvious way).  Continuing by induction, we find a sequence of points $x_k\in \Omega_k$ such that for each $k$, $\sum_j P^{k+1}(A_j(x_1,\dots,x_k))<1$.
Since $\bigcup_j A_j=\prod_k \Omega_k$, there must be some $j$ such that the sequence $x=(x_1,x_2,\dots)$ is in $A_j$.  But if $n$ is such that $E_j^{(k)}=\Omega_k$ for all $k>n$, then $A_j(x_1,\dots,x_n)=\prod_{k=n+1}^\infty \Omega_k$, so $P^{n+1}(A_j(x_1,\dots,x_n))=1$.  This is a contradiction, since the $x_i$ were chosen such that $\sum_j P^{n+1}(A_j(x_1,\dots,x_n))<1$.
